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Answer :
To find the velocity of the boat after Batman lands in it, we can use the principle of conservation of momentum. This principle states that the total momentum of a system remains constant if no external forces act on it.
Here is a step-by-step solution:
1. Identify the Known Values:
- Mass of Batman, [tex]\( m_{\text{Batman}} = 99.8 \, \text{kg} \)[/tex]
- Mass of the boat, [tex]\( m_{\text{boat}} = 445 \, \text{kg} \)[/tex]
- Initial velocity of the boat, [tex]\( v_{\text{initial}} = 13.5 \, \text{m/s} \)[/tex]
2. Determine the Total Mass After Batman Lands:
- When Batman lands in the boat, the total mass of the system (Batman plus the boat) will be:
[tex]\[
\text{Total mass} = m_{\text{Batman}} + m_{\text{boat}} = 99.8 \, \text{kg} + 445 \, \text{kg} = 544.8 \, \text{kg}
\][/tex]
3. Calculate the Initial Momentum:
- The momentum of the boat before Batman lands in it is given by:
[tex]\[
\text{Initial momentum} = m_{\text{boat}} \times v_{\text{initial}} = 445 \, \text{kg} \times 13.5 \, \text{m/s} = 6007.5 \, \text{kg} \cdot \text{m/s}
\][/tex]
4. Apply the Conservation of Momentum:
- Since Batman jumps straight down without any horizontal velocity, only the boat’s initial momentum is considered. The total momentum before Batman lands remains the same as after he lands:
[tex]\[
\text{Initial momentum} = \text{Final momentum}
\][/tex]
5. Determine the Final Velocity:
- Use the conservation of momentum to find the final velocity of the boat after Batman lands:
[tex]\[
\text{Final momentum} = \text{Total mass} \times \text{Final velocity}
\][/tex]
[tex]\[
6007.5 \, \text{kg} \cdot \text{m/s} = 544.8 \, \text{kg} \times \text{Final velocity}
\][/tex]
[tex]\[
\text{Final velocity} = \frac{6007.5 \, \text{kg} \cdot \text{m/s}}{544.8 \, \text{kg}} \approx 11.03 \, \text{m/s}
\][/tex]
Therefore, the velocity of the boat after Batman lands in it is approximately [tex]\( 11.03 \, \text{m/s} \)[/tex].
Here is a step-by-step solution:
1. Identify the Known Values:
- Mass of Batman, [tex]\( m_{\text{Batman}} = 99.8 \, \text{kg} \)[/tex]
- Mass of the boat, [tex]\( m_{\text{boat}} = 445 \, \text{kg} \)[/tex]
- Initial velocity of the boat, [tex]\( v_{\text{initial}} = 13.5 \, \text{m/s} \)[/tex]
2. Determine the Total Mass After Batman Lands:
- When Batman lands in the boat, the total mass of the system (Batman plus the boat) will be:
[tex]\[
\text{Total mass} = m_{\text{Batman}} + m_{\text{boat}} = 99.8 \, \text{kg} + 445 \, \text{kg} = 544.8 \, \text{kg}
\][/tex]
3. Calculate the Initial Momentum:
- The momentum of the boat before Batman lands in it is given by:
[tex]\[
\text{Initial momentum} = m_{\text{boat}} \times v_{\text{initial}} = 445 \, \text{kg} \times 13.5 \, \text{m/s} = 6007.5 \, \text{kg} \cdot \text{m/s}
\][/tex]
4. Apply the Conservation of Momentum:
- Since Batman jumps straight down without any horizontal velocity, only the boat’s initial momentum is considered. The total momentum before Batman lands remains the same as after he lands:
[tex]\[
\text{Initial momentum} = \text{Final momentum}
\][/tex]
5. Determine the Final Velocity:
- Use the conservation of momentum to find the final velocity of the boat after Batman lands:
[tex]\[
\text{Final momentum} = \text{Total mass} \times \text{Final velocity}
\][/tex]
[tex]\[
6007.5 \, \text{kg} \cdot \text{m/s} = 544.8 \, \text{kg} \times \text{Final velocity}
\][/tex]
[tex]\[
\text{Final velocity} = \frac{6007.5 \, \text{kg} \cdot \text{m/s}}{544.8 \, \text{kg}} \approx 11.03 \, \text{m/s}
\][/tex]
Therefore, the velocity of the boat after Batman lands in it is approximately [tex]\( 11.03 \, \text{m/s} \)[/tex].
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